Choosing balls out of a bag, replacing them by opposite colour Can somebody help me out on this question:

A bag contains four white balls and four red balls. A ball is selected at random, removed and replaced by a ball of the opposite colour. A second ball is then selected at random.
a) calculate the probability that the second ball was white.

Ok this is what i did but i'am not sure if its correct, i used the conditional probability formula, and got $(\frac{4}{8}\times\frac{\frac{3}{8}}{\frac{4}{8}})$
Thanks.
 A: Since the problem is symmetrical between red and white it is easy to see that the probability that the second ball is red will be the same as for the second ball being white. The probability will therefore be $\frac 12$.
A: If the first ball is white, the second will be white only with probaility $\frac38$.
If the first ball is red, the second will be white with probaility $\frac58$.
The two possibilities for the first ball each have probability $\frac48=\frac12$, so the answer is $\frac12\frac38+\frac12\frac58=\frac12$ (as one might have guessed immediately by symmetry).
A: Absolutely right mathematically, yet absolutely wrong, too. 
The options are either $\frac{3}{8}$ (half the time) or $\frac{5}{8}$ (half the time) which averages to $\frac{4}{8}$ but never is $\frac{4}{8}$ as such, (just like the "average family with $2.4$ children" does not exist.). Missing a target below and above in equal proportion is not the same thing as hitting it.
A: Conditional probability introduces knowledge about the result. Such a question would be what is the probability of the first ball being red given the second one is white?
Here, it's a straight forward approach:
before the first ball was drawn, the probability is: $P(Ball_{drawn_{1}}=white)=\frac{4}{8}=\frac{1}{2}$. After that, we can distinguish between two cases.I)White ball was drawn first $\Rightarrow$ $P(Ball_{drawn_{2}}=white | Ball_{drawn_{1}}=white )=\frac{3}{8}$, since one white ball previously was swapped to a red one.
II)Red ball was drawn first $\Rightarrow$ $P(Ball_{drawn_{2}}=white | Ball_{drawn_{1}}=red)=\frac{5}{8}$, since one red ball previously was swapped to a white one.
Now we remember that both cases happen with same probability ($P(Ball_{drawn_{1}}=white)=\frac{1}{2}$) and therefore  both cases account for half the solution, which is:
$$P(Ball_{drawn_{2}}=white)=\frac{1}{2}*P(Ball_{drawn_{2}}=white | Ball_{drawn_{1}}=white ) + \frac{1}{2}P(Ball_{drawn_{2}}=white | Ball_{drawn_{1}}=red)$$
I'm leaving the last step for you :)
